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  "body": "\n## Statement\n\nLet $k \\ge 1$ be an integer.\nThere is a constant $c_k$, depending only on $k$, with the following property.\nLet $a \u003c b$ be real numbers, let $\\phi \\colon [a, b] \\to \\R$ be a [[smooth function|0V33BNKZ9S]], and assume that $\\abs{\\phi^{(k)}(x)} \\ge 1$ for all $x \\in [a, b]$.\nIf $k = 1$, assume also that $\\phi'$ is monotonic on $[a, b]$.\nThen, for every $\\lambda \u003e 0$,\n$$\\Big| \\int_a^b e^{i\\lambda\\phi(x)} \\, dx \\Big| \\le c_k \\lambda^{-1/k}.$$\n\n## Remarks\n\nThe point is that the bound does not depend on $a$, $b$ or $\\phi$.\nFor $k = 1$, write $e^{i\\lambda\\phi} = (i\\lambda\\phi')^{-1} \\frac{d}{dx} e^{i\\lambda\\phi}$ and integrate by parts; monotonicity of $\\phi'$ makes the integral of $\\abs{(1/\\phi')'}$ telescope, and one gets $c_1 = 3$.\nFor $k \\ge 2$, use induction on $k$.\nOn the set where $\\abs{\\phi^{(k-1)}} \\ge \\lambda^{-1/k}$, apply the case $k - 1$ after rescaling.\nThis set misses at most an interval of length about $\\lambda^{-1/k}$, where the integrand is bounded by $1$.\n\nExample: for $\\phi(x) = x^2/2$ and $k = 2$, the lemma gives $\\abs{\\int_a^b e^{i\\lambda x^2/2} \\, dx} \\le c_2 \\lambda^{-1/2}$ uniformly in $a, b$; the Fresnel integral $\\int_{-\\infty}^{\\infty} e^{i\\lambda x^2/2} \\, dx = \\sqrt{2\\pi/\\lambda} \\, e^{i\\pi/4}$ shows that the power $\\lambda^{-1/2}$ cannot be improved.\n\nFor $k = 1$ the monotonicity of $\\phi'$ is needed.\nWithout it, no bound $c_1\\lambda^{-1}$ holds uniformly in $\\phi$ and in the length of the interval.\n\nA corollary adds an amplitude: if $\\psi$ is smooth on $[a, b]$, then $\\abs{\\int_a^b e^{i\\lambda\\phi}\\psi} \\le c_k\\lambda^{-1/k} \\big(\\abs{\\psi(b)} + \\int_a^b \\abs{\\psi'}\\big)$.\n\n## Explanations\n\n### L1 human\n\nAn integral of $e^{i\\lambda\\phi}$ is small when the phase $\\phi$ changes fast, because the integrand spins around the circle and cancels itself.\nThe lemma measures this: if some derivative of $\\phi$ of order $k$ is at least $1$, the integral is at most a constant times $\\lambda^{-1/k}$.\nHigher $k$ means a flatter possible phase and a weaker decay.\n\n### L2 human\n\nThis is the basic one-dimensional tool for [[oscillatory integrals|D5HZKE8E1A]].\nIt gives decay rates for Fourier transforms of measures on curves, such as $\\abs{\\hat\\sigma(\\xi)} \\lesssim \\abs{\\xi}^{-1/2}$ for arc length on a curve with nonzero curvature.\nIn higher dimensions there is no equally clean analogue, and one uses stationary phase or sublevel set estimates instead.\n",
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